Using standard methods will facilitate the planning and calculation of foundations, an example of calculating the foundation will simplify the calculation. Based on the recommendations given in the article, errors can be avoided during the construction of the selected structure (columnar, pile, tape or plate type).
Column base
For example, we use a one-story building with parameters in the plan of 6x6 m, as well as with walls of timber 15x15 cm (volumetric weight is 789 kg / m³), trimmed with clapboard lining on the outside. The basement of the building is made of concrete: height - 800 mm and width - 200 mm (bulk density of concrete materials - 2099 kg / m³). It is based on a reinforced concrete beam with a section of 20x15 (volume indicators of reinforced concrete - 2399). The walls are 300 cm high, and the slate roof has two slopes. The basement and the attic are made of boards located on beams with a section of 15x5, and are also insulated with mineral wool (insulation bulk density is 299 kg).
Knowing the load standards (according to SNiP), it is possible to correctly calculate the foundations. An example of calculating the foundation will allow you to quickly make calculations for your own building.
Load standards
- On the basement - 149.5 kg / m².
- In the attic - 75.
- The norm of snow load for terrain in the middle zone of the Russian Federation is 99 kg / m² relative to the roof area (in horizontal section).
- Different loads are pressured on different axes on the bases.
Pressure on each axis
Accurate indicators of structural and regulatory loads allow you to correctly calculate the foundations. An example of calculating the foundation is provided for the convenience of novice builders.
Design pressure along the axis "1" and "3" (extreme walls):
- From the log house of the ceiling: 600 x 300 cm = 1800 cm². This indicator is multiplied by the thickness of the vertical overlap of 20 cm (taking into account the external finish). It turns out: 360 cm³ x 799 kg / m³ = 0.28 t.
- From the rand beam: 20 x 15 x 600 = 1800 cm³ x 2399 ~ 430 kg.
- From the base: 20 x 80 x 600 = 960 cm³ x 2099 ~ 2160 kg.
- From the cap. The total mass of the entire overlap is calculated, then 1/4 of it is taken.
Logs with sides 5x15 are placed every 500 mm. Their mass is 200 cm³ x 800 kg / m³ = 1600 kg.
It is necessary to determine the mass of flooring and filing included in the calculation of foundations. An example of calculating the foundation indicates a insulation layer 3 cm thick.
The volume is 6 mm x 360 cm² = 2160 cm³. Further, the value is multiplied by 800, the total will be 1700 kg.
Mineral wool insulation is 15 cm thick.
Volume indicators are 15 x 360 = 540 cm³. When multiplied by a density of 300.01, we get 1620 kg.
Total: 1600.0 + 1700.0 + 1600.0 = 4900.0 kg. Divide everything by 4, we get 1.25 tons.
- From the attic ~ 1200 kg;
- From the roof: the total mass of one slope (1/2 roof), taking into account the mass of rafters, gratings and slate flooring - only 50 kg / m² x 24 = 1200 kg.
The load norm for columnar structures (for the axis "1" and "3" you need to find 1/4 of the total pressure on the roof) allows you to calculate the pile foundation. An example of the construction in question is ideal for printed construction.
- From the base: (600.0 x 600.0) / 4 = 900.0 x 150.0 kg / m² = 1350.0 kg.
- From the attic: 2 times less than from the basement.
- From snow: (100 kg / m² x 360 cm²) / 2 = 1800 kg.
As a result: the total indicator of structural loads is 9.2 tons, standard pressure - 4.1. For each axis "1" and "3" there is a load of about 13.3 tons.
Design pressure along the axis "2" (middle longitudinal line):
- From a frame of wall ceilings, a rand beam and a basement surface, the loads are similar to the values of the axis “1” and “3”: 3000 + 500 + 2000 = 5500 kg.
- From the basement and the attic, they have double indicators: 2600 +2400 = 5000 kg.
Below is the normative load and the calculation of the foundation foundation. An example is used in approximate values:
- From the cap: 2800 kg.
- From the attic: 1400.
As a result: the total design pressure is 10.5 tons, the standard load is 4.2 tons. The “2” axis has a weight of about 14,700 kg.
Axial pressure “A” and “B” (transverse lines)
Calculations are made taking into account the structural weight from the log of wall ceilings, rand beams and a socle (3, 0.5 and 2 tons). The pressure on the foundation along these walls will be: 3000 + 500 +2000 = 5500 kg.
Number of posts
To determine the required number of poles with a cross section of 0.3 m, soil resistance (R) is taken into account:
- With R = 2.50 kg / cm² (a frequently used indicator) and a shoe shoe area of 7.06 m² (for simplicity of calculations, take a lower value of 7 m²), the bearing capacity index of one column will be: P = 2.5 x 7 = 1 , 75 t.
- An example of calculating the columnar foundation for soil with resistance R = 1.50 takes the following form: P = 1.5 x 7 = 1.05.
- At R = 1.0, one pillar is characterized by a bearing capacity of P = 1.0 x 7 = 0.7.
- The resistance of watery soil is 2 times less than the minimum values of the tabulated indicators of 1.0 kg / cm². At a depth of 150 cm, the average is 0.55. The bearing capacity of the column is P = 0.6 x 7 = 0.42.
For the selected house, a volume of 0.02 m³ of reinforced concrete is required.
Placement points
- For wall ceilings: along the lines "1" and "3" with a weight of ~ 13.3 tons.
- On the axis "2" with a weight of ~ 14700 kg.
- Under the wall ceilings along the axes “A” and “B” with a weight of ~ 5500 kg.
If you need a calculation of the foundation for tipping, an example of calculations and formulas are given for large cottages. For suburban areas, they are not used. Particular attention is paid to load balancing, which requires careful calculation of the number of columns.
Examples of calculating the number of columns for all types of soil
Example 1:
R = 2.50 kg / cm²
For wall ceilings along the segment "1" and "3":
13.3 / 1.75 ~ 8 pillars.
On the axis "2":
14.7 / 1.75 ~ 9 pcs.
On segments "A" and "B":
5.5 / 1.75 = 3.1.
A total of approximately 31 posts. The volumetric index of concrete material is 31 x 2 mm³ = 62 cm³.
Example 2:
R = 1.50
On the line "1" and "3" ~ 12 columns.
On the axis "2" ~ 14.
On segments "A" and "B" ~ 6 each.
Total ~ 50 pieces. Volumetric indicator of concrete material ~ 1.0 m³.
Example 3:
Below you can find out how the calculation of a monolithic foundation is carried out . An example is given for soil with a tabular indicator R = 1,0. It has the following form:
On the line "1" and "2" ~ 19 pcs.
On the wall "2" ~ 21.
On segments "A" and "B" ~ 8 each.
Total - 75 posts. The volumetric index of concrete material is ~ 1.50 m³.
Example 4:
R = 0.60
On the line "1" and "3" ~ 32 pcs.
On the axis "2" ~ 35.
On segments "A" and "B" ~ 13.
Total - 125 pillars. Volumetric index of concrete material ~ 250 cm³.
In the first two calculations, corner posts are set at the intersection of the axes, and along the longitudinal lines with the same pitch. Reinforced concrete randbalks are cast in the formwork under the basement on the column heads.
In example No. 3, 3 columns are placed on intersecting axes. A similar number of bases are grouped along the axes "1", "2" and "3". Among builders, this technology is called "bushes." On a separate “bush”, it is required to install a common reinforced concrete grill head with its further placement on poles located on the axes “A” and “B” of the beam.
Example 4 allows you to build “bushes” of 4 columns at the intersection and along the longitudinal part of the lines (1-3) with further installation of grill heads. On them are placed randbalki under the basement.
Tape base
For comparison, the strip foundation is calculated below. An example is given taking into account the depth of the trench 150 cm (width - 40). The canal will be filled with sand mixture of 50 cm, then it will be filled with concrete to a height of one meter. It will require the development of soil (1800 cm³), laying of sand fraction (600) and concrete mix (1200).
Of the 4-column bases, the third is taken for comparison.
Drill work is carried out on an area of 75 cm³ with soil utilization of 1.5 cubic meters, or 12 times less (the rest of the soil is used for backfilling). The need for concrete mix is 150 cm³, or 8 times less, and in the sand fraction - 100 (it is necessary under the supporting beam). An exploratory pit is created near the foundation, which allows you to find out the condition of the soil. According to table data 1 and 2, the resistance is selected.
Important! In the bottom lines, these data will allow calculating the slab foundation - an example is indicated for all types of soil.
Sand resistance
Tab. 1
The resistance of the soil to the base, kg / cm 3| Sand fraction | Density level |
| Dense | Medium density |
| Large | 4.49 | 3.49 |
| Average | 3.49 | 2.49 |
| Shallow: Wet / Wet | 3-2.49 | 2 |
| Dusty: Wet / Wet | 2.49-1.49 | 2-1 |
Tab. 2
Clay soil resistance| The soil | Level
porosity | Soil resistance
kg / cm 3 |
| Solid | Plastic |
| Sandy loam | 0.50 / 0.70 | 3.0-2.50 | 2.0-3.0 |
| Loam | 0.50-1.0 | 2.0-3.0 | 1.0-2.50 |
| Clay soil | 0.50-1.0 | 2,50-6,0 | 1.0-4.0 |
Slab foundation
At the first stage, the thickness of the plate is calculated. The combined mass of the room is taken, including the weight of the installation, cladding and additional loads. According to this indicator and the area of the slab in the plan, the pressure from the room to the soil without the weight of the base is calculated.
It is calculated how much slab mass is lacking for a given pressure on the soil (for fine sand this figure will be 0.35 kg / cm², average density 0.25, hard and plastic sandy loam 0.5, hard clay 0.5 and plastic clay 0.25).
The area of the foundation should not exceed the conditions:
S> Kh × F / Kp × R,
where S is the sole of the base;
Kh is the coefficient for determining the reliability of the support (it is 1.2);
F is the total weight of all plates;
Kp - coefficient determining the conditions of work;
R is the soil resistance.
Example:
- The free mass of the building is 270,000 kg.
- The parameters in the plan are 10x10, or 100 m².
- Soil - loam with a moisture content of 0.35 kg / cm².
- The density of reinforced concrete is 2.7 kg / cm³.
The weight of the slabs is 80 tons behind - that's 29 cubic meters of concrete mix. For 100 squares, its thickness corresponds to 29 cm, so 30 is taken.
The total mass of the slab is 2.7 x 30 = 81 tons;
The total mass of the building with the foundation is 351.
The plate has a thickness of 25 cm: its mass is 67.5 tons.
We get: 270 + 67.5 = 337.5 (soil pressure is 3.375 t / m²). This is enough for an aerated concrete house with a cement density of V22.5 compression (slab grade).
Definition of tipping structure
The moment MU is determined taking into account the wind speed and the area of the building on which the impact is carried out. Additional fastening is required if the following condition is not met:
MU = (Q - F) * 17.44
F - the lifting force of the wind on the roof (in the given example, it is 20.1 kN).
Q is the calculated minimum asymmetric load (by the condition of the problem, it is 2785.8 kPa).
When calculating the parameters, it is important to consider the location of the building, the presence of vegetation and the structures erected nearby. Much attention is paid to weather and geological factors.
The above indicators are used for clarity of work. If you need to build a building yourself, it is recommended to consult with specialists.